1. Let
• v₀ = initial velocity of the car
• a = 2.3 m/s², its acceleration
• x = 22.7 m, the distance covered by the car in 1.3 s
Since acceleration is constant, the distance covered after this time is
x = v₀t + 1/2 at ²
Solve for v₀ :
x = v₀t + 1/2 at ²
v₀t = x - 1/2 at ²
v₀ = x/t - 1/2 at
Plug in everything you know:
v₀ = (22.7 m)/(1.3 s) - 1/2 (2.3 m/s²) (1.3 s) ≈ 16 m/s
2. Recall that under constant acceleration,
v ² - v₀² = 2ax
where v is the velocity after some unknown time. This time,
• v₀ = 0 because the is stopped at the light
• a = 3.7 m/s²
• x = 47.0 m
Solve for v :
v ² - v₀² = 2ax
v ² = v₀² + 2ax
v = √(v₀² + 2ax)
Plug in everything you know:
v = √(0² + 2 (3.7 m/s²) (47.0 m)) ≈ 19 m/s
3. Use the same equation as in the previous problem, but this time you solve for a.
v ² - v₀² = 2ax
a = (v ² - v₀²)/(2x)
We're given
• v₀ = 23 m/s
• v = 0, because the boy wants to stop his bike
• x = 80 m
Then
a = (0² - (23 m/s)²)/(2 (80 m)) ≈ -3.3 m/s²
That is, the brakes have to apply a force that decelerates the bike with magnitude 3.3 m/s².
4. Same formula for acceleration as before,
a = (v ² - v₀²)/(2x)
with
• v = 17.3 m/s
• v₀ = 0
• x = 48.4 m
We end up with
a = ((17.3 m/s)² - 0²)/(2 (48.4 m)) ≈ 3.09 m/s²
5. By definition of average acceleration,
a (ave) = ∆v/∆t
That is, average acceleration is the ratio of change in velocity to change in time. Here we have
• ∆v = 0 - 38.0 m/s = -38.0 m/s, the difference between the final and initial velocities
• ∆t = 3.9 s, the time it takes to stop
Then
a (ave) = (-38.0 m/s)/(3.9 s) ≈ -9.7 m/s²
